2016 AMC 8

It is best to split hours and minutes into parts, one of hours and another of minutes. We know that there is minutes in a hour. Therefore, there are minutes in hours. Adding the second part(the minutes) we get .

Solution 1

Using the triangle area formula for triangles: where is the area, is the base, and is the height. This equation gives us .

Solution 2

A triangle with the same height and base as a rectangle is half of the rectangle's area. This means that a triangle with half of the base of the rectangle and also the same height means its area is one quarter of the rectangle's area. Therefore, we get .

Let be the remaining student's score. We know that the average, , is equal to . We can use basic algebra to solve for :

giving us the answer of .

Solution 1

When Cheenu was a boy, he could run miles in hours and minutes minutes minutes, thus running minutes per mile. Now that he is an old man, he can walk miles in hours minutes minutes, thus walking minutes per mile. Therefore, it takes him minutes longer to walk a mile now compared to when he was a boy.

Solution 2

From the question, the old man can travel five miles in two hours, so we can set both speeds to miles. We can see that Cheenu as an old man takes hours and minutes for him to travel miles, which is also minutes. We can then divide this by fifteen, which gives us , thus the answer is

Solution 1

From the second bullet point, we know that the second digit must be , for a number divisible by ends in zero. Since there is a remainder of when is divided by , the multiple of must end in a for it to have the desired remainder We now look for this one:

The number satisfies both conditions. We subtract the biggest multiple of less than to get the remainder. Thus, .

Solution 2

We know that the number has to be one more than a multiple of , because of the remainder of one, and the number has to be more than a multiple of , which means that it has to end in a . Now, if we just list the first few multiples of adding one to the number we get: . As we can see from these numbers, the only one that has a three in the units place is , thus we divide by , getting , hence, .

Solution 1

We first notice that the median name will be the name. The name is .

Solution 2

To find the median length of a name from a bar graph, we must add up the number of names. Doing so gives us . Thus the index of the median length would be the 10th name. Since there are names with length , and names with length , the th name would have letters. Thus our answer is .

Our answer must have an odd exponent in order for it to not be a square. Because is a perfect square, is also a perfect square, so our answer is .

Solution 1

We can group each subtracting pair together:

After subtracting, we have:

There are even numbers, therefore there are even pairs. Therefore the sum is

Solution 2

Since our list does not end with one, we divide every number by and we end up with

We can group each subtracting pair together:

There are now pairs of numbers, and the value of each pair is . This sum is . However, we divided by originally so we will multiply to get the final answer of

The prime factorization is . Since the problem is only asking us for the distinct prime factors, we have . Their desired sum is then .

Solution 1

Let us plug in into . Thus it would be . Now we have . Plugging into , we have . Solving for we have

Solution 2

Let us set a variable equal to . Solving for y in the equation , we see that y is equal to five. By substitution, we see that = 5. Solving for x in the equation we get

We can see that the original number can be written as , where represents the tens digit and represents the units digit. When this number is added to the number obtained by reversing its digits, which is , the sum would be . From this, we can construct the equation , which simplifies to . Since there are 7 pairs of such digits and , , the answer would be

Solution 1

Let there be boys and girls in the school. We see , which means kids went on the trip and kids are girls. So, the answer is , which is .

Solution 2

Using WLOG (Without loss of generativity), Let there be boys and girls in the school. Now we can do to get the total number of students going to the field trip to be . Since we already know the number of girls to be . We have our answer to be . So, the answer is .

Solution 1

. Identify the total number of ways to select two different numbers from the set:

The set has elements. The number of ways to choose different numbers from is given by the combination formula: .

. Identify the favorable outcomes:

For the product to be zero, one of the chosen numbers must be zero. The set contains one zero (). To have a product of zero, we need to choose and any other number from the remaining five numbers .

The number of ways to choose 0 and one other number from the remaining five is .

. Calculate the probability:

The probability is the number of favorable outcomes divided by the total number of outcomes: .

Thus, the probability that the product is is

Solution 2 

Because the only way the product of the two numbers is is if one of the numbers we choose is we calculate the probability of NOT choosing a We get Therefore our answer is

Since he uses a gallon of gas every miles, he had used gallons after miles. Therefore, after the first leg of his trip he had gallons of gas left. Then, he bought gallons of gas, which brought him up to gallons of gas in his gas tank. When he arrived, he had gallons of gas. So he used gallons of gas on the second leg of his trip. Therefore, the second part of his trip covered miles. Adding this to the miles, we see that he drove miles.

First, we use difference of squares on to get . Using difference of squares again and simplifying, we get . Realizing that we don't need the right-hand side because it doesn't contain any factor of , we see that the greatest power of that is a divisor is .

Solution 1

Each lap Bonnie runs, Annie runs another quarter lap, so Bonnie will run four laps before she is overtaken. This means that Annie and Bonnie are equal so that Annie needs to run another lap to overtake Bonnie. That means Annie will have run laps.

Solution 2

Saying that Bonnie runs meters per second, then Annie will run meters per second. It will take Bonnie seconds to run lap and it would take seconds for Annie to do lap. The LCM of and is . In seconds Bonnie will do laps and Annie will do laps. Since Annie did more lap she passed Bonnie. So the answer is laps.

Solution 1

For the first three digits, there are combinations since is not allowed. For the final digit, any of the numbers are allowed. .

Solution 2

Counting the prohibited cases, we find that there are 10 of them. This is because, when we start with 9,1, and 1, we can have any of the 10 digits for the last digit. So, our answer is

Solution 1

From any th race, only will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal. Starting with the first race:

Adding all of the numbers in the second column yields

Solution 2

Every race eliminates players. The winner is decided when there is only runner left. You can construct the equation: - = . Thus, players have to be eliminated. Therefore, we need games to decide the winner, or

Solution 1

Let be the consecutive even integer that's being added up. By doing this, we can see that the sum of all even numbers will simplify to since . Now, . Remembering that this is the integer, we wish to find the , which is .

Solution 2

Let be the smallest number. The equation will become, . After you combine like terms, you get which turns into . , so . Then, you add .

Solution 1

We wish to find possible values of , , and . By finding the greatest common factor of and , we can find that is 3. Moving on to and , in order to minimize them, we wish to find the least such that the least common multiple of and is , . Similarly, with and , we obtain . The least common multiple of and is

Solution 2

The factors of are . The factors of are . The numbers that repeat are and so either has to be or . If is then is and is and the least common multiple of and are . We don't have to test because is the lowest answer, so if b equaling resulted in the least common multiple being less that then the correct answer won't be there. So the answer is .

We put five chips randomly in order and then pick the chips from the left to the right. To find the number of ways to rearrange the three red chips and two green chips, we solve for . However, we notice that whenever the last chip we draw is red, we pick both green chips before we pick the last red chip. Similarly, when the last chip is green, we pick all three red chips before the last green chip. This means that the last chip must be green in all the situations that work. This means we are left with finding the number of ways to rearrange three red chips and one green chip, which is . Because a green chip will be last out of the situations, our answer is .

The area of trapezoid is . Next, we find the height of each triangle to calculate their area. The two non-colored isosceles triangles are similar, and are in a ratio by AA similarity (alternate interior and vertical angles) so the height of the larger is while the height of the smaller one is Thus, their areas are and . Subtracting these areas from the trapezoid, we get . Therefore, the answer to this problem is .

Observe that is equilateral (all are radii of congruent circles). Therefore, . Since is a straight line, we conclude that . Since (both are radii of the same circle), is isosceles, meaning that . Similarly, .

Now, . Therefore, the answer is .

Solution 1 

We see that since is divisible by , must equal either or , but it cannot equal , so . We notice that since must be even, must be either or . However, when , we see that , which cannot happen because and are already used up; so . This gives , meaning . Now, we see that could be either or , but is not divisible by , but is. This means that and .

Solution 2

We know that out of is divisible by . Therefore is obviously 5 because is divisible by 5. So we now have as our number. Next, let's move on to the second piece of information that was given to us. is divisible by 3. So, according to the divisibility by 3 rule, the sum of has to be a multiple of 3. The only 2 big enough are 9 and 12 and since 5 is already given. The possible sums of are 4 and 7. So, the possible values for are 1,3,4,3 and the possible values of are 3,1,3,4. So, using this we can move on to the fact that is divisible by 4. So, using that we know that has to be even so 4 is the only possible value for . Using that we also know that 3 is the only possible value for 3. So, we have = so the possible values are 1 and 2 for and . Using the divisibility rule of 4 we know that has to be divisible by 4. So, either 14 or 24 are the possibilities, and 24 is divisible by 4. So the only value left for is 1. .